Question

求解两题高数题。

Answer 1

解：1。用洛必达法则：
原式=lim（x→0）[4(x-sinx)]/[4(e^x-1)^3*e^x]
=lim（x→0）[x-(x-x^3/3!)]/[(1+x-1)^3(1+x)]
=lim（x→0）x^3/[3!*x^3(1+x)
=1/6.
2. ∫[sin(2x)/√(4-cos^4x)]dx
=∫sin(2x)/√[(2+cos^2x)(2-cos^2x)]dx
=∫sin(2x)/√[(5/2-1/2+cos^2x)(3/2+1/2-cos^2x)]dx
=2∫sin(2x)/√[(5-cos2x)(3-cos2x)]dx
=2∫sin(2x)dx/√[15-8cos(2x)+cos^2(2x)]
=∫d[4-cos(2x)]/√{[4-cos(2x)]^2-1}
=ln{4-cos(2x)+√{[4-cos(2x)]^2-1}}+C